In an equilateral triangle ABC, D is a point on the side BC such that BD = ^{1}/_{3} BC. Prove that 9AD^{2} = 7AB^{2}

### Solution

Construction: Draw AP ⟂ BC

In Δ ADP,

AD^{2} = AP^{2} + DP^{2}

AD^{2} = AP^{2} + (BP - BD)^{2}

AD^{2} = AP^{2} + BP^{2} + BD^{2} - 2(BP)(BD)

AD^{2} = AB^{2} + (^{1}/_{3} BC)^{2} - 2(BC/2)(BC/3)

AD^{2} = ^{7}/_{9} AB^{2} (Since BC = AB)

9AD^{2} = 7AB^{2}